Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.8 AIME Find the answer

If $n$ is a positive integer such that $n^{3}+2 n^{2}+9 n+8$ is the cube of an integer, find $n$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

Since $n^{3}<n^{3}+2 n^{2}+9 n+8<(n+2)^{3}$ , we must have $n^{3}+2 n^{2}+9 n+8=(n+1)^{3}$ . Thus $n^{2}=6 n+7$ , so $n=7$ .

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